m273q multivariable calculus spring 2017 review problems

M273Q Multivariable Calculus Spring 2017 Review Problems for Exam 1

The formulas in the box will be provided on the exam.

κ(s) =

∣∣∣∣∣∣∣∣dTds∣∣∣∣∣∣∣∣ κ(x) =

|f ′′(x)|[1 + (f ′(x))2]3/2

κ(t) =||T′(t)||||r′(t)||

κ(t) =||r′(t)× r′′(t)||||r′(t)||3

1. True or False? Circle ONE answer for each. Hint: For effective study, explain why if ‘true’ and give acounterexample if ‘false.’

(a) T or F: If a⊥b and b⊥c, then a⊥c.

Solution: False. Let a = i,b = j, and c =< 1, 0, 1 > . Then a · b = i · j = 0 and b · c =<0, 1, 0 > · < 1, 0, 1 >= 0. However, a · c =< 1, 0, 0 > · < 1, 0, 1 >= 1.

(b) T or F: If a · b = 0, then ||a× b|| = ||a||||b||.

Solution: True. Assume both a and b are nonzero. (If either is zero the result is obvious). If θis the angle between a and b, then θ = π

2 since a and b are orthogonal. ||a×b|| = ||a||||b|| sin θ =||a||||b|| sinπ/2 = ||a||||b||.

(c) T or F: For any vectors u,v in R3, ||u× v|| = ||v × u||.

Solution: True. If θ is the angle between u and v, then ||u×v|| = ||u||||v|| sin θ = ||v||||u|| sin θ =||v × u||. (Or, ||u× v|| = || − v × u|| = | − 1|||v × u|| = ||v × u||.)

(d) T or F: The vector < 3,−1, 2 > is parallel to the plane 6x− 2y + 4z = 1.

Solution: False. A normal vector to the plane is n =< 6,−2, 4 > . Because < 3,−1, 2 >= 12n,

the vector is parallel to n and hence perpendicular to the plane.

(e) T or F: If u · v = 0, then u = 0 or v = 0.

Solution: False. For example i · j = 0 but i 6= 0 and j 6= 0.

(f) T or F: If u× v = 0, then u = 0 or v = 0.

Solution: False. For example i× i = 0 but i 6= 0.

(g) T or F: If u · v = 0 and u× v = 0, then u = 0 or v = 0.

Solution: True. If both u and v are nonzero, then u · v = 0 implies u and v are orthogonal.But u×v = 0 implies that u and v are parallel. Two nonzero vectors can’t be both parallel andorthogonal, so at least one of them must be 0.

(h) T or F: The curve r(t) =⟨0, t2, 4t

⟩is a parabola.

Solution: True. Parametric equations for the curve are x = 0, y = t2, z = 4t, and since t = z4

we have y = t2 =(z

4

)2or y =

z2

16, x = 0. This is an equation of a parabola in the yz−plane.

M273Q Multivariable Calculus Review Problems for Exam 1 - of 9

(i) T or F: If κ(t) = 0 for all t, the curve is a straight line.

Solution: True. Notice that κ(t) = 0 ⇐⇒ ||T′(t)|| = 0 ⇐⇒ T′(t) = 0 for all t. But thenT(t) = C, a constant vector, which is true only for a straight line.

(j) T or F: Different parameterizations of the same curve result in identical tangent vectors at a givenpoint on the curve.

Solution: False. For example, r1(t) =< t, t > and r2(t) =< 2t, 2t > both represent the sameplane curve (the line y = x), but the tangent vector r1(t) =< 1, 1 > for all t, while r1(t) =<2, 2 > . In fact, different parametrizations give parallel tangent vectors at a point, but theirmagnitudes may differ.

2. Which of the following are vectors?

(a)√

Vector © Scalar © Nonsense [(a · b)c]× a

(b) © Vector © Scalar√

Nonsense c× [(a · b)× c]

(c)√

Vector © Scalar © Nonsense c× [(a · b)c]

(d) © Vector√

Scalar © Nonsense (a× b) · c

3. Which of the following are meaningful?

(a)√

Meaningful © Nonsense ||w||(u× v)

(b) © Meaningful√

Nonsense (u · v)×w

(c)√

Meaningful © Nonsense u · (v ×w)

4. Find the values of x such that the vectors < 3, 2, x > and < 2x, 4, x > are orthogonal.

Solution: For the two vectors to be orthogonal, we need < 3, 2, x > · < 2x, 4, x >= 0. That is,3(2x) + 2(4) + x(x) = 0, or x = −2 or x = −4.

5. Find the decomposition a = a‖b + a⊥b of a =< 1, 1, 1 > along b =< 2,−1,−3 > .

Solution: a‖b =a · bb · b

b =

⟨−2

7,

1

7,

3

7

⟩. Finally, a⊥b = a − a‖b =< 1, 1, 1 > −

⟨−2

7,

1

7,

3

7

⟩=⟨

9

7,

6

7,

4

7

⟩.

6. Let a =

⟨1√2,

1√2

⟩,b =

⟨1√2,−1√

2

⟩, and u =< 3, 0 > .

(a) Show that a and b are orthogonal unit vectors.

Solution: a · b = 0 and ||a|| =√12+12√

2= 1, and ||b|| =

√12+(−1)2√

2= 1.

(b) Find the decomposition of u along a.

Solution: u||a = (u · a)a =3√2

⟨1√2,

1√2

⟩=

⟨3

2,

3

2

⟩. Thus, u⊥a =< 3, 0 > −

⟨32 ,

32

⟩=⟨

32 ,−

32

⟩.

(c) Find the decomposition of u along b.


Solution: Similarly, u||b = (u · b)b =

(< 3, 0 > ·

⟨1√2,− 1√

2

⟩)⟨1√2,− 1√

2

⟩=

⟨3

2,−3

2

⟩.

Thus, u⊥b =< 3, 0 > −⟨32 ,

32

⟩=⟨32 ,−

32

⟩= u⊥a.

7. (a) Find an equation of the sphere that passes through the point (6, -2, 3) and has center (-1,2,1).

Solution: Use the distance formula to find the distance between (6, -2, 3) and (-1,2,1). Then,the equation for the circle is (x+ 1)2 + (y − 2)2 + (z − 1)2 = 69.

(b) Find the curve in which this sphere intersects the yz-plane.

Solution: The intersection of this sphere with the yz−plane is the set of points on the spherewhose x−coordinate is 0. Putting x=0 in to the equation, (y − 2)2 + (z − 1)2 = 68, whichrepresents a circle in the yz−plance with center (0, 2, 1) and radius

√68.

8. For each of the following quantities (cos θ, sin θ, x, y, z, and w) in the picture below, fill in the blank withthe number of the expression, taken from the list to the right, to which it is equal.

x

y

z

w

a

b

θ

cos θ = 5 1.a · b||a||

sin θ= 4 2.a · b||b||

x = 2 3.||a× b||||b||

y= 3 4.||a× b||||a||||b||

z= 7 5.|a · b|||a||||b||

w= 6 6. ||b− a||

7.(b− a) · b||b||

9. Find an equation for the line through (4,−1, 2) and (1, 1, 5).

Solution: The line has direction v =< −3, 2, 3 > . Letting P0 = (4,−1, 2), parametric equations are

x = 4− 3t, y = −1 + 2t, z = 2 + 3t.


10. Find an equation for the line through (−2, 2, 4) and perpendicular to the plane 2x− y + 5z = 12.

Solution: A direction vector for the line is a normal vector for the plane, n =< 2,−1, 5 >, andparametric equations for the line are

x = −2 + 2t, y = 2− t, z = 4 + 5t.

11. Find an equation of the plane through (2, 1, 0) and parallel to x+ 4y − 3z = 1.

Solution: Since the two planes are parallel, they will have the same normal vectors. Then we cantake n =< 1, 4,−3 > and an equation of the plane is 1(x− 2) + 4(y − 1)− 3(z − 0) = 0.

12. Find an equation of the plane that passes through the point (−1,−3, 2) and contains the line x(t) =−1− 2t, y(t) = 4t, z(t) = 2 + t.

Solution: To find an equation of the plane we must first find two nonparallel vectors in the plane,then their cross product will be a normal vector to the plane. But since the given line lies in theplane, its direction vector a =< −2, 4, 1 > is one vector parallel to the plane. To find another vectorb which lies in the plane pick any point on the line [say (−1, 0, 2), found by setting t = 0] and let b bethe vector connecting this point to the given point in the plane (−1,−3, 2). (But beware; we shouldfirst check that the given point is not on the given line. If it were on the line, the plane wouldnt beuniquely determined. What would n then be when we set n = a × b?) Thus b =< 0, 3, 0 > andn = a×b =< 0 + 3, 0− 0, 6− 0 >=< 3, 0, 6 > and an equation of the plane is 3(x+ 1) + 6(z− 2) = 0or x+ 2z = 3.

13. Find the point at which the line x(t) = 1− t, y = t, z(t) = 1 + t and the plane z = 1− 2x+ y intersect.

Solution: Substitute the line into the equation of the plane. z = 1−2x+y ⇒ 1+t = 1−2(1−t)+t⇒t = 1⇒ the point of intersection is (0, 1, 2).

14. (a) Find an equation of the plane that passes through the points A(2, 1, 1), B(−1,−1, 10), and C(1, 3,−4.)

Solution: The vector−−→AB =< −3,−2, 9 > and

−→AC =< −1, 2,−5 > lie in the plane, so n =−−→

AB ×−→AC =< −8,−24,−8 > or equivalently, < 1, 3, 1 > is a normal vector to the plane. The

point A(2, 1, 1) lies on the plane so an equation for the plane is 1(x−2) + 3(y−1) + 1(z−1) = 0.

(b) A second plane passes through (2, 0, 4) and has normal vector < 2,−4,−3 > . Find an equation forthe line of intersection of the two planes.

Solution: The point (2,0,4) lies on the second plane, but the point also satisfies the equationof the first plane, so the point lies on the line of intersection of the planes. A vector v in thedirection of this intersecting line is perpendicular to the normal vectors of both planes, so takev =< 1, 3, 1 > × < 2,−4,−3 >=< −5, 5,−10 > or just use < 1,−1, 2 > . Parametric equationsfor the line are

x = 2 + t, y = −t, z = 4 + 2t.

15. Provide a clear sketch of the following traces for the quadratic surface y =√x2 + z2 + 1 in the given

planes. Label your work appropriately.

x = 0;x = 1; y = 0; y = 2; z = 0.


x

y

x

z

y

z

__-axis

__-axis

__-axis

__-axis

Solution:

16. Match the equations with their graphs. Give reasons for your choices.

(a) II 8x+ 2y + 3z = 0

(b) I z = sinx+ cos y

(c) IV z = sin

(π

2 + x2 + y2

)(d) III z = ey

17. Find a vector function that represents the curve of intersection of the cylinder x2 + y2 = 16 and the planex+ z = 5.

Solution: The projection of the curve C of intersection onto the xy−plane is the circle x2 + y2 =16, z = 0. So we can write x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ 2π. From the equation of the plane, we havez = 5 − x = 5 − 4 cos t, so parametric equations for C are x = 4 cos t, y = 4 sin t, z = 5 − 4 cos t, 0 ≤t ≤ 2π, and the corresponding vector function is r(t) =< 4 cos t, 4 sin t, z = 5− 4 cos t >, 0 ≤ t ≤ 2π.

18. Find an equation for the tangent line to the curve x = 2 sin t, y = 2 sin 2t, and z = 2 sin 3t at the point(1,√

3, 2).


Solution: The curve is given by r(t) =< 2 sin t, 2 sin 2t, 2 sin 3t >, so r′(t) =< 2 cos t, 4 cos 2t, 6 cos 3t >. The point (1,

√3, 2) corresponds to t = π/6, so the tangent vector there is r′(π/6) =<

√3, 2, 0 > .

Then the tangent line has direction vector <√

3, 2, 0 > and includes the point (1,√

3, 2), so parametricequations are x = 1 +

√3t, y =

√3 + 2t, z = 2.

19. A helix circles the z−axis, going from (2, 0, 0) to (2, 0, 6π) in one turn.

(a) Parameterize this helix.

Solution: r(t) =< 2 cos t, 2 sin t, 3t > . (Note that 1 revolution is 2π, so 2π(3) = 6π.)

(b) Calculate the length of a single turn.

Solution: For 0 ≤ t ≤ 2π, ||r′(t)|| =√

4 + 9 =√

13. Thus s =

∫ 2π

0

√13dt = 13(2π).

(c) Find the curvature of this helix.

Solution: The unit tangent vector is T(t) = 1√13

< −2 sin t, 2 cos t, 3 >, so T′(t) = −2√13

<

cos t, sin t, 0 >. Thus, ||T′(t)|| = 2√13. Since ||r′(t)|| =

√13, κ = 2

13

20. (a) Sketch the curve with vector function r(t) = 〈t, cosπt, sinπt〉 , t ≥ 0.

Solution: The corresponding parametric equations for the curve are x = t, y = cosπt, z = sinπt.Since y2 + z2 = 1, the curve is contained in a circular cylinder with axis the x−axis. Since x = t,the curve is a helix.

(b) Find r′(t) and r′′(t).

Solution: Since r(t) = 〈t, cosπt, sinπt〉 , r′(t) = 〈1,−π sinπt, π cosπt〉 , and r′′(t) =⟨0,−pi2 cosπt,−π2 sinπt

⟩.

21. Which curve below is traced out by r(t) =

⟨sinπt, cosπt,

1

4t2⟩, 0 ≤ t ≤ 2.

Solution: Graph 1. Note that r(0) =< 0, 1, 0 > .

22. Find a point on the curve r(t) =⟨t+ 1, 2t2 − 2, 5

⟩where the tangent line is parallel to the plane x+ 2y−

4z = 5.


Solution: The plane has normal vector n =< 1, 2,−4 > . Since r′(t) =< 1, 4t, 0 >, we want <1, 4t, 0 > · < 1, 2,−4 >= 0. That is, 1 + 8t = 0, and r(− 1

8 ) =< 78 ,−6342 , 5 > .

23. Let r(t) =⟨√

2− t, (et − 1)/t, ln(t+ 1)⟩.

(a) Find the domain of r.

Solution: The expressions√

2− t, (et− 1)/t, and ln(t+ 1) are all defined when 2− t ≥ 0. Thus,t ≤ 2, t 6= 0, and t+ 1 > 0⇒ t > −1. Finally, the domain of r is (−1, 0) ∪ (0, 2].

(b) Find limt→0

r(t).

Solution: limt→0

r(t) =⟨√

2, 1, 0⟩. Note that in the y− component we use l’Hospital’s Rule.

(c) Find r′(t).

Solution: r′(t) =⟨− 1

2√2−t ,

tet−et+1t2 , 1

t+1

⟩.

24. Suppose that an object has velocity v(t) =⟨3√

1 + t, 2 sin(2t), 6e3t⟩

at time t, and position r(t) =<0, 1, 2 > at time t = 0. Find the position, r(t), of the object at time t.

Solution: r(t) =

∫v(t)dt =

⟨∫3√

1 + tdt,

∫2 sin(2t)dt,

∫6e3tdt

⟩=⟨

2(1 + t)3/2 + c1,− cos(2t) + c2, 2e3t + c3

⟩.

Thus, r(0) =< 2 + c1,−1 + c2, 2 + c3 >=< 0, 1, 2 >⇒ c1 = −2, c2 = 2, c3 = 0. So, r(t) =⟨2(1 + t)3/2 − 2,− cos(2t) + 2, 2e3t

⟩.

25. If r(t) =⟨t2, t cosπt, sinπt

⟩, evaluate

∫ 1

0

r(t)dt.

Solution:

∫ 1

0

r(t)dt =

⟨∫ 1

0

t2dt,

∫ 1

0

t cosπtdt,

∫ 1

0

sinπtdt

⟩=

⟨1

3,− 2

π2,

2

π

⟩.

26. Find the length of the curve: x = 2 cos(2t), y = 2t3/2, and z = 2 sin(2t); 0 ≤ t ≤ 1.

Solution: s =

∫ 1

0

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt =

∫ 1

0

√16 sin2(2t) + 16 cos2(2t) + 9t dt =

∫ 1

0

√16 + 9tdt =

122

27.

27. Reparameterize the curve r(t) =< et, et sin t, et cos t > with respect to arc length measured from the point(1, 0, 1) in the direction of increasing t.

Solution: The parametric value corresponding to the point (1, 0, 1) is t = 0. Since r′(t) =< et, et(cos t+

sin t), et(cos t−sin t) >, ||r′(t)|| = et√

1 + (cos t+ sin t)2 + (cos t− sin t)2 =√

3et, and s(t) =∫ t0eu√

3du =√

3(et−1)⇒ t = ln(

1 + 1√3s). Therefore, r(t(s)) =

⟨1 + 1√

3s,(

1 + 1√3s)

sin ln(

1 + 1√3s),(

1 + 1√3s)

cos ln(

1 + 1√3s)⟩

.


28. Find the tangent line to the curve of intersection of the cylinder x2 + y2 = 25 and the plane x = z at thepoint (3, 4, 3).

Solution: Let x(t) = 5 cos t, y(t) = 5 sin t, and z(t) = 5 cos t, so that r(t) =< 5 cos t, 5 sin t, 5 cos t >and r′(t) =< −5 sin t, 5 cos t,−5 sin t > . When (x, y, z) = (3, 4, 3), x(t) = z(t) = 5 cos t = 3, andy(t) = 5 sin t = 4, so r′(t0) =< −4, 3,−4 > is a direction vector for the line. The tangent line hasparametric equations x(t) = 3 − 4t, y(t) = 4 + 3t, and z(t) = 3 − 4t. ANOTHER SOLUTION: Let

x(t) = t, y =√

25− x2 =√

25− t2, and z(t) = x(t) = t. Then r(t) =⟨t, (25− t)1/2, t

⟩. In this

case, r′(t) =⟨

1, −t√25−t2 , 1

⟩. When x = 3, t = 3, and r′(3) =< 1, −34 , 1 > . So the tangent line has

parametric equations x(t) = 3 + t, y(t) = 4− 34 t, and z(t) = 3 + t.

29. For the curve given by r(t) =⟨13 t

3, t2, 2t⟩, find

(a) the unit tangent vector

Solution: T(t) =r′(t)

||r′(t)||=< t2, 2t, 2 >√t4 + t2 + 1

=< t2, 2t, 2 >

(t2 + 2).

(b) the unit normal vector

Solution:

T′(t) =< 4t, 4− 2t2,−4t >

(t2 + 2)2⇒ ||T′(t)|| =

√4(t4 + 4t2 + 4)

(t2 + 2)2=

2

t2 + 2and N(t) =

< 2t, 2− t2,−2t >

t2 + 2

(c) the curvature

Solution: κ(t) =||T′(t)||||r′(t)||

=2

(t2 + 2)2

30. A particle moves with position function r(t) =< t ln t, t, e−t >. Find the velocity, speed, and accelerationof the particle.

Solution: v(t) = r′(t) =< 1 + ln t, 1,−e−t > . ν(t) = ||v(t)|| =√

(1 + ln t)2 + 1 + (−e−t)2 =√2 + 2 ln t+ (ln t)2 + e−2t. a(t) = v′(t) =<

1

t, 0, e−t > .

31. A particle starts at the origin with initial velocity< 1,−1, 3 > and its acceleration is a(t) =< 6t, 12t2,−6t >. Find its position function.

Solution: v(t) =

∫a(t)dt =

⟨∫6t dt,

∫12t2 dt,

∫−6t dt

⟩=⟨3t2, 4t3,−3t2

⟩+C, but< 1,−1, 3 >=

v(0)+C, so C =< 1,−1, 3 > and v(t) =⟨3t2 + 1, 4t3 − 1,−3t2 + 3

⟩. r(t) =

∫v(t)dt =

⟨t3 + t, t4 − t, 3t− t3

⟩=

D. But r(0) = 0, so D = 0, and r(t) =⟨t3 + t, t4 − t, 3t− t3

⟩.

32. A flying squirrel has position r(t) =

⟨t+

t2

2, 1− t, 2 + t2

⟩at time t. Compute the following at time t = 1:

(a) The velocity at time t = 1, v(1) = < 2,−1, 2 >.


Solution: v(t) = r′(t) =< 1 + t,−1, 2t > .

(b) The speed at time t = 1, ν(1) = 3 .

Solution: ν = ||v(t)||. ||v(1)|| =√

4 + 1 + 4 = 3.

33. Consider the vector valued function r(t) = describing the curve shown below. Put the curvature of r atA,B and C in order from smallest to largest. Draw the osculating circles at those points.

Solution: B, A, C.

m273q multivariable calculus spring 2017 review problems

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